If ∑i=120 20Ci−1 20Ci+20Ci−13=k21, then k equals

A
50
B
200
C
100
D
400

Solution:

Given, $\sum_{i = 1}^{20} {(\frac{^{20} C_{i - 1}}{^{20} C_{i} +^{20} C_{i - 1}})}^{3} = \frac{k}{21}$

$\begin{array}{l} \Rightarrow \sum_{i = 1}^{20} {(\frac{^{20} C_{i - 1}}{^{21} C_{i}})}^{3} = \frac{k}{21} \Rightarrow \sum_{i = 1}^{20} {(\frac{i}{21})}^{3} = \frac{k}{21} \\ \Rightarrow \frac{1}{(21)^{3}} {[\frac{20 (21)}{2}]}^{2} = \frac{k}{21} \Rightarrow 100 = k \end{array}$

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